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Linear algebra projection: http://postimage.org/image/4n6220ftf/ (I've got the formulas and thought I understood the concept.. but am not coming to the answer the book has
 2 years ago
 2 years ago
Linear algebra projection: http://postimage.org/image/4n6220ftf/ (I've got the formulas and thought I understood the concept.. but am not coming to the answer the book has
 2 years ago
 2 years ago

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phiBest ResponseYou've already chosen the best response.1
let b= (1,1,1) let n be a scalar we can say b= nL + e where e is a vector orthogonal to L dw:1328148406952:dw
 2 years ago

phiBest ResponseYou've already chosen the best response.1
use L^T(b  nL)= 0 (L^T means L transpose. e and L perpendicular) to find the scalar n n = L^T b/(L^T L) to find the reflection of b = nL + e, go in the other direction b reflected = nL  e
 2 years ago

phiBest ResponseYou've already chosen the best response.1
e= b  nL so b_reflected = nL  b + nL = 2nL b
 2 years ago

CynosureEPRBest ResponseYou've already chosen the best response.0
Lemme give it a go
 2 years ago

CynosureEPRBest ResponseYou've already chosen the best response.0
Can you explain L^T? in attempting to find n?
 2 years ago

phiBest ResponseYou've already chosen the best response.1
the T means transpose. L^T L is just L dot L, and L^T b is L dot b (dot product)
 2 years ago

CynosureEPRBest ResponseYou've already chosen the best response.0
Okay.. so I'm extremely close. I've gotten (1/9) of: [11] [ 1 ] [11] The book has the same, but positive. I'm guessing my math is just off somewhere.
 2 years ago

phiBest ResponseYou've already chosen the best response.1
I get n = L dot b / (L dot L) = 5/9 and e = b  nL = (1,1,1)  (5/9) (2,1,2) = (1, 4, 1)/9 b_reflected = nL  e = (5/9) (2,1,2)  (1, 4, 1)/9 = (10,5,10)/9 + (1, 4, 1)/9 b_reflected = (11,1,11)/9 Here's a picture http://www.wolframalpha.com/input/?i=plot+vector+%281%2C1%2C1%29+%2C+vector+%281%2C2%2C1%29%2C++vector%2811%2F9%2C1%2F9%2C11%2F9%29%2C+vector+%2811%2F9%2C1%2F9%2C11%2F9%29
 2 years ago

CynosureEPRBest ResponseYou've already chosen the best response.0
Many, many thanks! I will attempt another problem using this method and see if I get the correct answer.
 2 years ago

CynosureEPRBest ResponseYou've already chosen the best response.0
Aha! You are a Godsend. <3
 2 years ago
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